这个要命的要求能不能用shell实现

关阴月飞

case when COL1 = 'V' AND COL2 = 'B' AND COL3 = 'C' AND P_CODE = 0 AND M_SRC = 5 AND M_FEC = 0 AND SEC_TYP = 1 THEN 869
WHEN COL1 = 'V' AND COL2 = 'B' AND P_CODE = 0 AND M_SRC = 5 AND M_FEC = 0  THEN 999
WHEN COL1 = 'V' AND COL2 = 'I' AND  P_CODE = 0 AND M_FEC = 0 THEN 873
WHEN COL1 = 'F' AND COL2 = 'B' AND P_CODE = 0 AND M_SRC = 5 AND M_FEC = 0 THEN 1233
END
上面这段是不是rule 0: 要的结果?

那rule 1  rule 2  rule 3 对应的结果是怎么样的呢?

sunzhiguolu

perl abc.pl a b






#!/usr/bin/perl
use strict;
use warnings;

sub Load{
    local @ARGV = pop;
    my $fb = [];
    while(<>){
        my @aT = split;
        /\A\s*[a-z]/i ? push(@{$fb}, [@aT]) : push(@{$fb->[-1]}, @aT ? @aT : undef);
    }
    $fb;
}

my $fb = Load(pop);
my ($pos, $fa) = (-1, []);

print("case ");
<>;
my @aFix = ('COL1',  'COL2',  'COL3',  'P_CODE',  'M_SRC',  'M_FEC',  'SEC_TYP');
local $" = " AND ";
while(<>){
    if(/\A(\d+)\Z/){
        $pos = $1;
        next;
    }
    next if(!defined($fb->[$pos]));
    print("when ");
    my @aT = split;
    my @aOut = ();
    foreach(0 .. $#aT - 1){
        next if(!defined($fb->[$pos][$_]));
        my $v = $fb->[$pos][$_];
        my $pair = $aT[$_] eq 'D' ? [$aFix[$_], 0] : [$aFix[$_], $v =~ /\D/ ? "'$v'" : $v];
        push(@aOut, sprintf("%s = %s", @$pair));
    }
    print("@aOut THEN $fb->[$pos][-1]\n");
}
print("END\n");